populating-next-right-pointers-in-each-node

题目描述

Follow up for problem “Populating Next Right Pointers in Each Node”.
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
You may only use constant extra space.

For example,
Given the following binary tree,
1
/ \
2 3
/ \ \
4 5 7

After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ \
4-> 5 -> 7 -> NULL

如果借助辅助空间可以直接按照层序遍历来做,如果不借助辅助空间可以利用next进行遍历,下面的代码有优化的空间.

/** * Definition for binary tree with next pointer. * struct TreeLinkNode { *  int val; *  TreeLinkNode *left, *right, *next; *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */class Solution {public:    void connect(TreeLinkNode *root) {        if(root==NULL)            return;        while(root!=NULL) //遍历完所有的层结束            {            TreeLinkNode* temp=root;                        while(root!=NULL){     //遍历完一层结束                if(root->left&&root->right)  //如果左右孩子都有将左右孩子连起来                {                root->left->next=root->right;                          }                if(root->left!=NULL&&root->right==NULL)  //只有左孩子的情况                {                    TreeLinkNode* temp2=root->next;                    while(temp2){                       //一直到找到可以连接的下一个点位置                    if(temp2->left){                        root->left->next=temp2->left;                        break;                    }                    if(temp2->right){                        root->left->next=temp2->right;                        break;                    }                    temp2=temp2->next;                    }                    root=temp2;                       //root更新为已连接的孩子的父节点                    continue;                                    }                if(root->right!=NULL)                 //有右孩子的情况包括左右孩子都有                {                    TreeLinkNode* temp2=root->next;                    while(temp2){                      //寻找下一个连接的点.                    if(temp2->left)                    {                    root->right->next=temp2->left;                    break;                    }                    if(temp2->right){                    root->right->next=temp2->right;                    break;                    }                    temp2=temp2->next;                    }                    root=temp2;                    continue;                }                root=root->next;                    //左右孩子都没有直接更新到右边一个节点            }            while(temp)                              //找到下一层最左边的节点,继续遍历下一层.            {                if(temp->left){                    temp=temp->left;                    break;                }                else if(temp->right)                    {                    temp=temp->right;                    break;                }                temp=temp->next;            }            root=temp;        }    }};